b)
I assume bigger the angle A, smaller is F[v1]. Why?
I assume it's because F[v] is constant no matter what the angle A is, but why is that?

2)
According to my book angles A and A1 are the same:

Code (Text):

[B]F[net] = m * g * sin[A1] = m * g * sin[A][/B].

I'd imagine angle A being the same as angle A1 only if F[g] = F[v1]. Then direction of F[net] would be horizontal. But since that is not the case thus the two angles shouldn't be the same.

3)
I will quote my book:

Distance of a ball from equilibrium state can be stated with

Code (Text):

[B]L = A1 * d = A * d [/B]

, where L is arc of a circle. When at angle A, the net force on the ball is F = m * g * sin[A], which gives the ball acceleration

Code (Text):

[B]a = - g * sin[A][/B]

Acceleration vector a certainly isn't linear with L = A * d, and thus the osciliation isn't harmonic. But it becomes harmonic, if angle A is small enough for us to replace sin[A] with A “

a)

a certainly isn't linear with L = A * d

I assume by that they mean to say that when arc L is twice as great, a isn't twice as great.
But what has that got to do with harmonic osiclation? Is with harmonic oscillation a linear with L?
Can you show me some proof of that?

b)

But it becomes harmonic, if angle A1 is small enough for us to replace sin[A1] with A1

First of all, I'm not sure that sin[A1] and A1 are ever roughly the same size, since no matter how small A1 is, sin[A1] will always be 100 or more times smaller. Right?

c)
Second, even if sin[A1] and A1 have about the same value when A1 is small enough, what is the purpose of replacing sin[A1] with A1? Why do we want to do that?

d)
Also, why is acceleration vector a negative?
I realise that when a has opposite direction to ball's velocity that it has to be negative. But sometimes ball's velocity and acceleration vectors have same direction and thus a should be positive?

Um if your angle keeps increasing it eventually reaches 90, so it keeps getting smaller and when it reaches 90 it goes to zero and tension is only determined in x direction since your tension component is basically Tension*cos(theta). I hope that answered the angle part of your problem